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FIBONACCI NUMBERS AND TRIGONOMETRIC IDENTITIES
N. GARNIER AND O. RAMARÉ
Abstract. In 1969 Webb and Parberry proved a startling trigonometric identity involving
Fibonacci numbers. This identity has remained isolated up to now, despite the amount of
work on related polynomials. We provide a wide generalization of this identity together
with what we believe (and hope!) to be its proper understanding.
1. Introduction
Fibonacci numbers verify a wealth of identities, see e.g. [5], [6], [7], [12]. By specifying x
and y to 1 in Corollary 10 of [4], we get an intriguing one which states that for n ≥ 1:
¶ [(n−1)/2]
¶
[(n−1)/2] µ
Y
Y µ
2kπ
2 kπ
Fn =
=
.
(1)
1 + 4 cos
3 + 2 cos
n
n
k=1
k=1
Webb and Parberry’s paper [15] contains all the necessary material to write this identity, but
they do not state it explicitly. This formula is indeed intriguing: the left hand side satisfies
a second order recursion formula while no such recursion arises from the right hand side
2kπ
expression. Indeed, how could we connect cos 2kπ
and cos n+1
? Taking a number theoretic
n
point
√ of view leads to more dismay. Fibonacci numbers are linked with the arithmetic of
Q( 5) and not with that of Q(exp(2iπ/n)).
The mystery gets somewhat lifted by the proof of the above identity. We introduce the
second order Chebyshev polynomials Un by
U0 (x) = 1,
U1 (x) = x,
Un+1 (x) = 2xUn (x) − Un−1 (x)
(2)
so that they verify
sin(n + 1)θ
.
sin θ
This last expression leads to the recursion above as well as to the formula
¶
n µ
Y
kπ
n
Un (x) = 2
x − cos
.
n+1
k=1
Un (cos θ) =
(3)
(4)
Since it is not difficult to discover that Fn = in−1 Un−1 (−i/2) we get the result. This factorization has been noted and studied by [1], [4], [15], and [16]. The main arguments of this
proof are the recurrence relation (2)
√ and the rule of additions of sine. However this sheds no
light on how the arithmetic of Q( 5) and Q(exp(2iπ/n)) get entangled.
Elaborating on this argument, Webb & Parberry in [15] and Hoggatt & Long in [4] defined
a sequence of polynomials by
u0 (x, y) = 0,
56
u1 (x, y) = 1,
un+2 (x, y) = xun+1 (x, y) + yun (x, y).
(5)
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FIBONACCI NUMBERS AND TRIGONOMETRIC IDENTITIES
The case y = 1, the only one considered in [15], leads to what is sometimes called Fibonacci
polynomials. A further generalization led to Morgan-Voyce polynomials in [8], [13] and [14],
and more recently to Brahmagupta’s polynomials in [10] and [11]. In particular, it is proved
that
¶
n µ
Y
kπ
√
un (x, y) =
x − 2i y cos
(6)
n
k=1
and this leads to amazing identities like the following.
Corollary 1. We have for n ≥ 3
¶
Y µ
2 2π`
1 + 4 sin
= (1 + Fn − 2Fn+1 + (−1)n )2
n
1≤`≤n
where Fn is the nth Fibonacci number, with F0 = 0 and F1 = 1.
We also have 1 + Fn − 2Fn+1 + (−1)n = 1 − Ln + (−1)n where Ln is the nth Lucas number
(Ln = Fn−1 + Fn+1 ).
This corollary does not appear as such in [4], but can be derived from the material presented therein. We provide a simple proof later. Again the LHS satisfies a linear recursion, which we now compute explicitly. First express Fn in terms of φ and −1/φ where φ,
the golden ratio, is the larger root of x2 − x − 1 and expands the square. There comes
a linear combination of terms of the form pn , where all the p’s appearing are roots of
(x2 − 3x + 1)(x2 − x − 1)(x + 1)(x − 1) = x6 − 4x5 + 2x4 + 6x3 − 4x2 − 2x + 1 since
φ2 and 1/φ2 are the roots of x2 − 3x + 1. If we call g(n) the LHS of the quantity computed
in the above corollary, we have
g(n + 6) = 2g(n + 5) + 4g(n + 4) − 6g(n + 3) − 2g(n + 2) + 4g(n + 1) − g(n)
since each sequence (pn ) satisfies it.
The next question is whether one can obtain such identities with three homogeneous
variables x, y and z. One path to such a generalization has been to study the dynamics of
the zeros of Fibonacci polynomials in [3], getting interesting by-products but no trigonometric
identity.
2. Results and Proofs
We prove here such a trigonometric identity. Let us start with a simple case.
Theorem 1. We have for n ≥ 3 and ξ = exp(2iπ/n)
Y ¡
¢
1 − xξ ` − yξ 2` = 1 − Gn + (−y)n
1≤`≤n
where Gn is defined by the recursion Gn = xGn−1 + yGn−2 for n ≥ 2 with G0 = 2, G1 = x
and G2 = x2 + 2y.
The polynomials Gn are a special case of Brahmagupta polynomials (take t = 1) introduced
and studied in [10], [11]. These are linked with Morgan-Voyce polynomials, see [8] and are
the Fibonacci polynomials when we further specialize y to be equal to 1.
At this level, our proof was a mystery to us, we did in fact stumble on it while studying
a completely different problem. It was from then onwards tempting to prove such trigonometric relations with r terms instead of 3. This led us to a very simple and direct proof of a
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THE FIBONACCI QUARTERLY
general relation, via considerations of circulant matrices. Most earlier connections between
determinants and Fibonacci numbers, or Chebyshev polynomials involved continuants. Determinant of a family of increasing size tridiagonal matrices whose first elements are properly
chosen while the later ones are fixed (see [1], [2], and exercise 24 page 85 of [5]).
And while studying this proof, we discovered that the heart of such identities was a hundred
years old one liner (see [9], paragraph 136, formula (4)).
Fundamental Lemma. Let (xs )0≤s≤r be complex numbers and set ξ = exp (2iπ/n). We
have
Y X
Y
(ρn − 1)
xs ξ s` = (−1)rn xnr
where P (Y ) =
Pr
s=0
0≤`≤n−1 0≤s≤r
ρ/P (ρ)=0
xs Y s .
Simply because the quantity to be computed is up to the sign of the resultant of P (Y )
and Qn (Y ) = Y n − 1 for which we have
Y
Y
Res(P, Qn ) = (−1)rn
P (u) = xnr
Qn (ρ).
(7)
u/Qn (u)=0
ρ/P (ρ)=0
So, on one side, we have the roots of one polynomial, while on the other one we have the
roots of another polynomial. This identity being so fundamental, we recall its one line proof.
Consider two polynomials A and B with respective leading terms am and bn , degrees m and
n and roots (αi )1≤i≤m and (βj )1≤j≤n repeated with multiplicity. Then, and it can be taken
as a definition,
Y
Res(A, B) = anm bm
(αi − βj ),
(8)
n
1≤i≤m
1≤j≤n
from which (7) follows trivially. There is an expression of this resultant as a (Sylvester)
determinant whose entries are the coefficients of A and B, as well as some 0’s, but we will
not invoke such an expression.
2
Let us use our fundamental
√ lemma on an example. Taking P = Y − Y − 1, with roots
the golden ratio φ = (1 + 5)/2 and −1/φ we get
¶
Y µ
2π`
1 + 2i sin
= 1 − φn − (−φ)−n + (−1)n
n
0≤`≤n−1
= 1 + Fn − 2Fn+1 + (−1)n
since φn = φ−1 Fn + Fn+1 and (−φ)−n = −φFn + Fn+1 , yielding an illuminating proof of
Corollary 1!
Proving (1) along these lines requires some tuning: we start from the polynomial P =
1 + iY + Y 2 and use only even indices 2n. This leads after some manipulations to the square
of the right-hand side of (1). We simplify the squareroot of the left-hand side by noting that
2F2n+1 − F2n − 2(−1)n = 5Fn2 .
(9)
We saw in these examples how linear recurrent sequences can be introduced and we now
address the general case treated in the Theorem. The adaptation is straightforward. To
avoid denominators, we set P (Y ) = Y 2 − xY − y with roots ρ1 and ρ2 in C(x, y). By our
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VOLUME 46/47, NUMBER 1
FIBONACCI NUMBERS AND TRIGONOMETRIC IDENTITIES
fundamental lemma, we have
Y ¡
¢
ξ 2` − xξ ` − y = (ρ1 ρ2 )n − ρn1 − ρn2 + 1
1≤`≤n
= (−y)n − ρn1 − ρn2 + 1.
We set Gn (x, y) = ρn1 + ρn2 , which yields G0 = 2, G1 = x and Gn+2 = xGn+1 + yGn . We need
only to factor ξ 2` out and exchange ` by −` to get our Theorem!
Let us now treat theP
general case. Let x0 , . . . , xr be r indeterminates. Let ρ1 , . . . , ρr be
the r roots of P (Y ) = 0≤s≤r xs Y s in an algebraic closure of K(x0 , . . . , xr ). We set
Y
(10)
(ρns − 1)
Hn = (−1)rn xnr
1≤s≤r
which in fact belongs to K[x0 , . . . , xr ] and
X
X X
Y
H (X) =
Hn X n =
(−1)r−|S|
ρns ((−1)r xr X)n
n≥0
=
X
S⊂{1,...,r}
n≥0 S⊂{1,...,r}
(11)
s∈S
(−1)r−|S|
Q
.
1 − (−1)r s∈S ρs xr X
(12)
This shows that H (X) is a rational fraction with denominator of degree at most 2r . As
a consequence Hn verifies a linear recursion of degree at most 2r which we easily establish
by writing H (X) = A (X)/B(X) with A (X) and B(X) polynomials. When equating
coefficients in the equation
X
B(X)
Hn X n = A (X)
n≥0
we recover an explicit form of the required recursion. Summarizing, we get
Theorem 2. Let (xs )0≤s≤r be complex numbers and set ξ = exp (2iπ/n). We have
Y X
xs ξ s` = Hn
where P (Y ) =
most 2r .
Pr
s=0
0≤`≤n−1 0≤s≤r
xs Y s and Hn defined by (10) satisfies a linear recursion of degree at
Since B(X) is fairly universal and determines the coefficients of the recursion satisfied by
Hn , it would be satisfactory to have a complete description of it solely in terms of the xs ’s.
We have not been able
Q to derive such a description. We can of course group together the
r
products 1 − (−1) s∈S ρs xr X over S with a fixed cardinality. As symmetric expressions
of the roots, each can be expressed as a polynomial of C[x0 , x1 , . . . , xr ]. For |S| = 0, we get
1 − (−1)r xr X. For |S| = 1, we get (−xr X)r P ((−1)r /(xr X)).
We end this section with yet another identity.
Corollary 2. We have for n ≥ 5
¶ (
[(n−4)/4] µ
Y
2kπ
Fn
5 + 4 sin2
= p 2
n
Fn − 4
k=1
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when n is even,
when n is odd.
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THE FIBONACCI QUARTERLY
We achieve this by considering the polynomial P = Y 2 − 3Y + 1 with roots φ2 and φ−2 .
We then follow the path described above together with some symmetry and identity (9).
Here are the details to the proof.
¶
n−1
n−1
Y¡
Y n−1
Yµ
¢
2πk
k
2k
k
1 − 3ξ + ξ
=
ξ
−3 + 2 cos
n
k=0
k=0
k=0
¶
n−1
Yµ
2πk
n−1
= (−1)
−3 + 2 cos
.
n
k=0
¶ (
(n−1)/2 µ
Y
2πk
−5 if n is even,
= (−1)n
−3 + 2 cos
×
n
1
if n is odd.
k=1
3. Extensions and Limitations
The above approach would work if we were to replace Y n − 1 by Q(Y n ) for a fixed
polynomial Q. It would also extend to the case when the coefficients of Q are polynomials in
n. The same remark holds for the coefficients of P . In these cases, the roots do not depend
on roots of unity, which means that the rule of addition of sine does not in fact intervene.
We can even handle the more difficult case Qn = Y 2n + Y n+1 + 1.
2
However, we are not able to treat the case Y n − 1 or any other non subsequence of Y n − 1
where the exponents would not be taken in an arithmetic progression. No linear recursion
in that case exists, for it would mean a linear recursion for the values Hn2 , and their growth
is too steep to allow such a fact (once we specialize the xs ’s).
Let us end this paper with a related problem. Restricting the product to indices ` prime
to n in our theorems, one gets the norm of 1 − xξ − xξ 2 in Q(ξ):
Corollary 3. We have
Y ¡
¢µ(n/d)
¢ Y¡
1 − Gd + (−y)d
1 − xξ ` − yξ 2` =
1≤`≤n
(`,n)=1
d|n
where µ denotes the Möbius function.
The question arose to decide whether this norm verifies a linear recursion as above or not.
Our approach via a resultant supports a negative answer. We have
Y ¡
Y
¢
Φn (ρ)
1 − xξ ` − yξ 2` = (−1)(r−1)n y n
1≤`≤n
(`,n)=1
ρ/P (ρ)=0
where Φn is the nth cyclotomic polynomial. In general, this polynomial does not have a
finite number of monomes (for instance when n is prime), which ruins the approach we used,
but also make us believe no recursion does indeed exist.
References
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Numbers, Fibonacci Q., 41.1 (2003), 13–19, 2003.
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Fibonacci Q., 42.3 (2004), 216–221.
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VOLUME 46/47, NUMBER 1
FIBONACCI NUMBERS AND TRIGONOMETRIC IDENTITIES
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MSC2000: 11B39
Université Lille 1, 59 655 Villeneuve d’Ascq, France
Laboratoire Paul Painlevé, Université Lille 1, 59 655 Villeneuve d’Ascq, France
E-mail address: [email protected]
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61